• Answer of this question

    Find four number in ap  whose sum is 20 and the sum of whose square is
    Davneet Singh's image
    Davneet Singh

    Let the numbers be a-2d, a-d, a and a+d.

     

    Their sum is

    a-2d+ a-d+ a + a+d = 20

    4a-2d = 20

    2a - d = 10 …(1)

     

    Now, Sum of square is

    (a - 2d)+ (a - d)+ a2 + (a + d)2 = 180

    a– 4ad + 4d^2 + a^2 – 2ad + d^2 + a^2 + a^2 + 2ad + d^2 = 180

    a^2 – 4ad + 4d^2 + a^2 + d^2 + a^2 + a^2 + d^2 = 180

    4a2 – 4ad + 6d2 = 180

    2a^2 – 2ad + 3d^2 = 90 …(2)

     

    From (1)

    d = 2a -10.

    Put that in (2) to get

    2a^2 -2a(2a–10) + 3(2a–10)^2 = 90, or

    2a^2 + 20a - 4a^2 + 3(4a^2 - 40a + 100) = 90

    2a^2+20a-4a^2+300–120a+12a^2 = 90

    10a^2–100a+210 = 0

    a^2–10a+21=0

    (a-3)(a-7) = 0

    Hence a = 3 or 7 and the corresponding value of d from (1)

    d=2a - 10 is -4 or 4.

     

    So the 4 terms of the AP are 3-[2*(-4)] = 11, 7, 3 and -1, or

    -1, 3, 7 and 11.

     

    Check:

    11 + 7 + 3 – 1 = 20. Correct.

    1^2+3^2+7^2+11^2 = 1+9+49+121 = 180. Correct.


    Written on June 10, 2018, 1:43 p.m.